{"id":325,"date":"2021-02-25T15:49:32","date_gmt":"2021-02-25T14:49:32","guid":{"rendered":"https:\/\/logbooks.ifosim.org\/pykat\/?p=325"},"modified":"2021-03-04T11:38:39","modified_gmt":"2021-03-04T10:38:39","slug":"shift-and-phase-change-from-tilted-beamsplitters-part-1-theory","status":"publish","type":"post","link":"https:\/\/logbooks.ifosim.org\/pykat\/blog\/shift-and-phase-change-from-tilted-beamsplitters-part-1-theory\/","title":{"rendered":"Shift and phase change from tilted beamsplitters – part 1: theory"},"content":{"rendered":"\n

Introduction<\/h1>\n\n\n\n

Given a z-shaped telescope setup with two beamsplitters such as depicted in Fig. 1, we calculate the effects on the outgoing beam when applying a small additional tilt \u03b2<\/em> to the angle of incidence \u03b1<\/em> of both beamsplitters.<\/p>\n\n\n\n

\"\"
Figure 1 – basic setup<\/figcaption><\/figure>\n\n\n\n

Such a small extra tilt is applied in Finesse 2 via the xbeta<\/strong><\/em> parameter of a bs<\/strong><\/em> component. With respect to the “untilted beam” (i.e. \u03b2<\/em> = 0) we calculate – for the outgoing beam – both the shift in its position \u2206<\/em> and the change of its phase \u03b4\u03c6<\/em>, each as a function of the angle \u03b2<\/em> and the distance s<\/em> between the beamsplitters. For the shift we derive eq. (5) below<\/p>\n\n\n\n

\\Delta = s \\cdot \\sin 2\\beta<\/div>\n\n\n\n

while for the phase change we find eq. (13) below<\/p>\n\n\n\n

\\delta \\varphi = \\frac{2 s \\cdot \\sin^2 \\beta}{\\lambda} \\cdot 360\\degree<\/div>\n\n\n\n

Note that this is the first post in a series, the second post (part 2: finesse examples<\/a>) will compare the results obtained here with simulations from finesse 2.<\/p>\n\n\n\n

Detailed Setup<\/h1>\n\n\n\n
\"\"
Figure 2 – Two beamsplitters, small tilt \u03b2<\/em><\/figcaption><\/figure>\n\n\n\n

Fig. 2 depicts the two different trajectories. The distance between the two beamsplitters is s<\/em>. The initial angle of incidence of the “untilted” beamsplitters (dark bue) is \u03b1<\/em> and its corresponding trajectory is plotted in dark red. The two beamsplitters are rotated over a small additional angle \u03b2<\/em> (light blue) with corresponding trajectory in light red.
As a result, the path traveled by the beam between the first and second beamsplitter changes from s<\/em> to s\u0303<\/em>; it will hit the second mirror at a different point (a distance d<\/em> away from the original point) and the outgoing beam will shift over a distance \u0394<\/em>. The difference in total path length between the dark and light red paths will cause a phase shift \u03b4\u03c6<\/em>. In the next subsections we will derive \u0394<\/em> and followed by \u03b4\u03c6<\/em>.<\/p>\n\n\n\n

Calculation of shift in position \u0394<\/h2>\n\n\n\n

Using the sine rule for the triangle enclosed by d<\/em>, s<\/em> and s\u0303<\/em><\/p>\n\n\n\n

\\frac{d}{\\sin 2\\beta} = \\frac{s}{\\sin \\gamma} = \\frac{\\tilde{s}}{\\sin(90\\degree+\\alpha-\\beta)} \\qquad (1)<\/div>\n\n\n\n

we can get the length d<\/em><\/p>\n\n\n\n

d = s \\cdot \\frac{\\sin 2\\beta}{\\sin \\gamma} \\qquad (2)<\/div>\n\n\n\n

Furthermore from the total angle of that same triangle we can get the angle \u03b3<\/p>\n\n\n\n

(90\\degree-\\alpha-\\beta+2\\alpha) + 2\\beta + \\gamma = 180\\degree \\quad \\to \\quad \\gamma = 90\\degree – \\alpha – \\beta \\qquad (3)<\/div>\n\n\n\n

We need to calculate \u0394<\/em>:<\/p>\n\n\n\n

\\Delta = d \\cdot \\sin (90\\degree-\\alpha-\\beta) = d \\cdot \\sin \\gamma \\qquad (4)<\/div>\n\n\n\n

where we used eq. (3). Using eq. (2) we then obtain the shift<\/p>\n\n\n\n

\\Delta = s \\cdot \\sin 2\\beta \\qquad (5)<\/div>\n\n\n\n

We see that all \u03b1 dependence cancels out. Furthermore, we see that the shift increases linearly with the distance between the two beamsplitters. Since for larger values of \u03b2<\/em> the paraxial approximation requires more and more higher modes to be included in a simulation, increasing s<\/em> instead of \u03b2<\/em> could help simulating the same shift \u0394<\/em> (we will look into convergence in the second post of this series).<\/p>\n\n\n\n

Calculation of difference in path length and \u03b4\u03c6<\/em><\/h2>\n\n\n\n

Note that there are two parts:<\/p>\n\n\n\n

  1. p1<\/sub>: the difference between s\u0303<\/em> and s<\/em><\/li>
  2. p2<\/sub>: horizontal part since we hit the beamsplitter further to the right<\/li><\/ol>\n\n\n\n

    p1<\/sub> will have a positive contribution, while p2<\/sub> will have a negative contribution to the change in path length.<\/p>\n\n\n\n

    calculation p1<\/sub><\/h3>\n\n\n\n

    First note that<\/p>\n\n\n\n

    \\sin \\gamma = \\sin(90\\degree – \\alpha – \\beta) = \\cos(\\alpha + \\beta) \\qquad (6)<\/div>\n\n\n\n

    We can obtain s\u0303<\/em> from the sine rule eq. (1) and use (6) to get<\/p>\n\n\n\n

    \\tilde{s} = s \\cdot \\frac{\\sin(90\\degree + \\alpha – \\beta)}{\\sin \\gamma}\n= s \\cdot \\frac{\\cos(\\alpha – \\beta)}{\\cos(\\alpha + \\beta)} \\qquad (7)<\/div>\n\n\n\n

    so<\/p>\n\n\n\n

    p_1 = \\tilde{s} – s\n= s \\cdot \\frac{\\cos(\\alpha – \\beta) – \\cos(\\alpha + \\beta)}{\\cos(\\alpha + \\beta)} \\qquad (8)<\/div>\n\n\n\n

    calculation p2<\/sub><\/h3>\n\n\n\n

    Since part p2<\/sub> shortens the total path, we add an overall minus sign:<\/p>\n\n\n\n

    p_2 = – d \\cdot \\cos(90\\degree-\\alpha-\\beta) = -d \\cdot \\sin(\\alpha + \\beta) \\qquad (9)<\/div>\n\n\n\n

    Using eq. (2) and (6) this becomes<\/p>\n\n\n\n

    p_2 = – s \\cdot \\frac{\\sin (\\alpha + \\beta) \\sin 2\\beta}{\\cos(\\alpha + \\beta)}\n \\qquad (10)<\/div>\n\n\n\n

    combining the two<\/h3>\n\n\n\n

    Combining eq. (8) and (10) we get<\/p>\n\n\n\n

    s \\cdot \\frac{\\cos (\\alpha – \\beta) – \\cos (\\alpha + \\beta) – \\sin (\\alpha + \\beta) \\sin 2\\beta}{\\cos (\\alpha + \\beta)} \\qquad (11)<\/div>\n\n\n\n

    We work out the numerator:<\/p>\n\n\n\n

    \n
    \\cos (\\alpha – \\beta) – \\cos (\\alpha + \\beta) – \\sin (\\alpha + \\beta) \\sin 2\\beta =<\/div>\n<\/div><\/div>\n\n\n\n
    \\cos\\alpha \\cos\\beta + \\sin\\alpha \\sin\\beta \n – \\cos\\alpha \\cos\\beta + \\sin\\alpha \\sin\\beta\n – (\\sin\\alpha \\cos\\beta + \\cos\\alpha \\sin\\beta) 2 \\sin\\beta \\cos\\beta =<\/div>\n\n\n\n
    2 \\sin\\alpha \\sin\\beta -2 (\\sin\\alpha \\sin\\beta \\cos^2\\beta + \\cos\\alpha \\cos\\beta \\sin^2\\beta) =<\/div>\n\n\n\n
    2 \\sin\\alpha \\sin\\beta ( 1 – \\cos^2\\beta) – 2 \\cos\\alpha \\cos\\beta \\sin^2\\beta =<\/div>\n\n\n\n
    2 (\\sin\\alpha \\sin\\beta – \\cos\\alpha \\cos\\beta) \\sin^2\\beta =<\/div>\n\n\n\n
    -2 \\cos(\\alpha + \\beta) \\sin^2 \\beta<\/div>\n\n\n\n

    Putting this into eq. (11) we see that the cos(\u03b1<\/em>+\u03b2<\/em>) factor cancels out and we find the total path difference<\/p>\n\n\n\n

    p_1 + p_2 = -2 s \\cdot \\sin^2 \\beta \\qquad (12)<\/div>\n\n\n\n

    Note that again the \u03b1<\/em> dependence cancels out. The phase shift resulting from the path difference can be calculated by dividing eq. (12) by the wavelength \u03bb and multiplying by 360\u00b0. We also pick up an extra<\/em> overall minus sign, cancelling the one from eq. (12), since a plane wave is defined as cos(\u03c9 t<\/em> – k x<\/em> + \u03c6<\/em>), see for example Bond et al. (2016) eq. (1.4). Putting it together we get<\/p>\n\n\n\n

    <\/div>\n\n\n\n
    \\delta \\varphi = \\frac{2 s \\cdot \\sin^2 \\beta}{\\lambda} \\cdot 360\\degree \\qquad (13)<\/div>\n\n\n\n

    Conclusions<\/h2>\n\n\n\n
    • Both the shift of the beam eq. (5) and the change in phase eq. (13) are in<\/em>dependent of the angle \u03b1<\/em><\/li>
    • Given the square in eq. (12) the path length difference is small but not per se compared to the wavelength \u03bb<\/em>, leading to a significant phase shift.<\/li><\/ul>\n\n\n\n

      Effect finite size beam<\/h1>\n\n\n\n

      The above calculation was done for an infinitesimally narrow beam hitting the beamsplitter exactly in the midpoint (its rotation point). We show here that the result is the same for a beam hitting the beamsplitter off-axis.<\/p>\n\n\n\n

      \"\"
      Figure 3 – Moving the original beamsplitter and original beam left<\/figcaption><\/figure>\n\n\n\n

      From Fig. 3 we see that – compared to the on-axis case – the effect of such an off-axis incoming beam can be described by a horizontal translation of the original beamsplitters (dark blue) with<\/em> their beams (dark red) with respect to the original tilted beamsplitters (light blue) and their beams (light red). It is clear that<\/p>\n\n\n\n

      • this does not affect the shift \u2206<\/em> of the beam on the beam detector.<\/li>
      • this does not affect the total path length difference and hence \u03b4\u03c6<\/em> (the dark red beam gains a bit on the left and loses the same amount on the right).<\/li><\/ul>\n","protected":false},"excerpt":{"rendered":"

        Introduction Given a z-shaped telescope setup with two beamsplitters such as depicted in Fig. 1, we calculate the effects on the outgoing beam when applying a small additional tilt \u03b2 to the angle of incidence \u03b1 of both beamsplitters. Such a small extra tilt is applied in Finesse 2 via the xbeta parameter of a […]<\/p>\n","protected":false},"author":40,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"ssl_alp_hide_revisions":true,"footnotes":"","ssl_alp_hide_crossreferences_to":false},"categories":[1],"tags":[],"ssl-alp-coauthor":[48],"class_list":["post-325","post","type-post","status-publish","format-standard","hentry","category-uncategorised"],"_links":{"self":[{"href":"https:\/\/logbooks.ifosim.org\/pykat\/wp-json\/wp\/v2\/posts\/325","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/logbooks.ifosim.org\/pykat\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/logbooks.ifosim.org\/pykat\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/logbooks.ifosim.org\/pykat\/wp-json\/wp\/v2\/users\/40"}],"replies":[{"embeddable":true,"href":"https:\/\/logbooks.ifosim.org\/pykat\/wp-json\/wp\/v2\/comments?post=325"}],"version-history":[{"count":35,"href":"https:\/\/logbooks.ifosim.org\/pykat\/wp-json\/wp\/v2\/posts\/325\/revisions"}],"predecessor-version":[{"id":452,"href":"https:\/\/logbooks.ifosim.org\/pykat\/wp-json\/wp\/v2\/posts\/325\/revisions\/452"}],"wp:attachment":[{"href":"https:\/\/logbooks.ifosim.org\/pykat\/wp-json\/wp\/v2\/media?parent=325"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/logbooks.ifosim.org\/pykat\/wp-json\/wp\/v2\/categories?post=325"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/logbooks.ifosim.org\/pykat\/wp-json\/wp\/v2\/tags?post=325"},{"taxonomy":"ssl-alp-coauthor","embeddable":true,"href":"https:\/\/logbooks.ifosim.org\/pykat\/wp-json\/wp\/v2\/ssl-alp-coauthor?post=325"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}